Culegere de probleme de Analiz˘a numeric˘a

138 Aproximarea funct¸ionalelor liniare
Solut¸ie.a = x0, xi = x0 +ih, i = 0,m, xm = b
ϕm+1(x) = h m+1
ϕm+1(x) =
m
i=0
m
(x−xi)
i=0
x = x0 +th
(t−i) = h m+1 ψm+1(t) = h m+1 t [m+1]
Lema 9.3.11 a) ϕm+1(xm/2 +σ) = (−1) m+1 ϕm+1(xm/2 −σ) unde xm
2 = x0 +
m
2 h.
b) De asemenea pentrua < σ +h < xm
2
s¸i pentruxm
2
Demonstrat¸ie.
ψm+1
ψm+1
< σ < b, σ = xi,
ψm+1
ψm+1
s¸i σ = xi
|ϕm+1(σ +h)| < |ϕm+1(σ)|
|ϕm+1(σ)| < |ϕm+1(σ +h)|
ψm+1(t) = t [m+1]
m
2 −s
m
= ψm+1
2 +s
pentrumimpar
m
2 −s
m
= −ψm+1
2 +s
pentru m par
m
2 −s
m
=
2 −s
m
2 −s−1
m
...
2 −s−m
m
2 +s
m
=
2 +s
m
2 +s−1
m
...
2 +s−m
ϕm+1(xm
2 +σ) = hm+1
m
ψ
= (2s+m)(2s+m−2)...(2s−m)
2 m
(9.5) ⇒ (2s−m)(2s−m+2)...(2s+m)
2m (−1) m+1
2 +σ
= (−1) m+1 h m+1
m
ψ
2 −σ
b)0